I need help with number 119

Rebecca had 9/10 of a yard of ribbon. Then she gave 7/20 of a yard to a friend. How much does she have left? Explain your work

lianna [129]

9/10 - 7/20=(9x20) - (7x10)= 10x20= 110/200 110 divided by 10/ 200 divided by 10= 11/20. So that brings you to your answer of 11/20. hope this helps :)

4 0

3 years ago

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Which of the equations will be a true statement if p=10 power of 3? Select all that apply.

Marat540 [252]

Answer:

A. 3.4 divided by p equals 0.034

B. 437 divided by p equals 0.437

C. 53.45 divided by p equals 53.45

D. 6,340 divided by p equals 6.34

E. 2,458.2 divided by p equals 24.582

Which of the equations will be a true statement if p=10 power of 3? Select all that apply.

The answer is A D and E

Step-by-step explanation:

7 0

3 years ago

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A nutrition company is marketing a low calorie snack brownie. A serving size of the snack is 3 brownies and has a total of 50 ca

lozanna [386]

Answer:

350 calories

Step-by-step explanation:

divide 50 by 3 then multiply that number by 21.

3 0

3 years ago

Are these functional or non functional ?

svetoff [14.1K]

Answer:

Relation 3 is a function and Relation 4 is nog

6 0

3 years ago

The U.S. and many other countries have recently tightened airport security. A sociologist is interested in estimating the propor

melisa1 [442]

Answer:

0.0073 < 0.05, which means that we reject the null hypothesis and conclude that the true proportion of interest is higher than 0.7.

Step-by-step explanation:

Conduct a test to determine whether the true proportion of interest is higher than 0.7.

This means that the null hypothesis is: $H_{0}: p = 0.7$

And the alternate hypothesis is: $H_{a}: p > 0.7$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.7 is tested at the null hypothesis:

This means that $\mu = 0.7, \sigma = \sqrt{0.7*0.3}$

The sociologist found that 375 of the 500 travelers randomly selected and interviewed indicated that the airports were safe.

This means that $n = 500, X = \frac{375}{500} = 0.75$

Value of the z-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.75 - 0.7}{\frac{\sqrt{0.7*0.3}}{\sqrt{350}}}$

$z = 2.44$

P-value of the test:

Probability of z being larger than 2.44, that is, a proportion larger than 0.75.

This is, looking at the z-table, 1 subtracted by the pvalue of Z = 2.44. S

Z = 2.44 has a pvalue of 0.9927

1 - 0.9927 = 0.0073

0.0073 < 0.05, which means that we reject the null hypothesis and conclude that the true proportion of interest is higher than 0.7.

8 0

3 years ago